Gil Kalai’s blog

Diameter Problem (4)

Let us consider another strategy to deal with our diameter problem. Let us try to associate other graphs to our family of sets.

Recall that we consider a family of subsets of size of the set .

Let us now associate more general graphs to as follows: For an integer define as follows: The vertices of are simply the sets in . Two vertices and are adjacent if . Our original problem dealt with the case . Thus, . Barnette proof presented in the previous part refers to and to paths in this graph.

As before for a subset let denote the subfamily of all subsets of which contain . Of course, the smaller is the more edges you have in . It is easy to see that assuming that is connected for every for which is not empty already implies our condition that is connected for every for which is not empty.

Let be the maximum diameter of in terms of and , for all families of -subsets of satisfying our connectivity relations.